Question

The distance of the midpoint of the line joining two points $\left(4,0\right)$ and $\left(0,4\right)$ from the centre of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=16$ is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $3\sqrt{2}$
• D. $2\sqrt{3}$

Answer 1

The correct option is B

$2\sqrt{2}$

Explanation for the correct option:

Finding the distance of the midpoint of the line from the centre of the circle:

Given, ${\mathrm{x}}^2+{\mathrm{y}}^2=16$

The two points joining the midpoint are $\left(0,4\right)$ and $\left(4,0\right)$.

Using the mid-point formula,

$\mathrm{Midpoint}=\left(\frac{4+0}{2},\frac{0+4}{2}\right)$

$=\left(2,2\right)$

And Center $=\left(0,0\right)$.

Using the distance formula,

$\begin{array}{rcl}\text{Distance}& =& \sqrt{{\left(x_1-x\right)}^2+{\left(y_1-y\right)}^2}\\ & =& \sqrt{{\left(2-0\right)}^2+{\left(2-0\right)}^2}\end{array}$

$$\begin{gathered} =\sqrt{4+4} \\ =\sqrt{8} \\ =\sqrt{4\times 2} \\ =2\sqrt{2} \end{gathered}$$

Hence, option (B) is the correct answer.